A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field `vecE` at the centre O is

Options

A. `(lamda)/(2pi in_(0) a^(2))`
B. `(lamda)/(4 in_(0) a)`
C. `(lamda^(2))/(2pi in_(0) a)`
D. `(lamda)/(2pi in_(0) a)`

Answer

D

Explanation

Electric intensity at centre `0`, due to small element `dl` of charged ring
`dE = (lambda dl)/(4pi in_(0) a^(2)) = (lambda (a d theta))/(4 pi in_(0) a^(2))`
As is clear from Fig horizontal components of `dE` will cancel out in pairs and vertical components wil add

`:. E = int_(0)^(pi) dE sin theta = int_(0)^(pi) (lambda)/(4 pi in_(0) a) sin theta d theta`
`= (lambda)/(4 pi in_(0)a) [- cos theta]_(0)^(pi)`
`E = (-lambda)/(4pi in_(0) a) (cos pi - cos 0) = (lambda)/(2 pii in_(0) a)`